### (0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

minus(minus(x)) → x
minus(h(x)) → h(minus(x))
minus(f(x, y)) → f(minus(y), minus(x))

Rewrite Strategy: INNERMOST

### (1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS to CDT

### (2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(minus(z0)) → z0
minus(h(z0)) → h(minus(z0))
minus(f(z0, z1)) → f(minus(z1), minus(z0))
Tuples:

MINUS(minus(z0)) → c
MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
S tuples:

MINUS(minus(z0)) → c
MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS

Compound Symbols:

c, c1, c2

### (3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

Removed 1 trailing nodes:

MINUS(minus(z0)) → c

### (4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(minus(z0)) → z0
minus(h(z0)) → h(minus(z0))
minus(f(z0, z1)) → f(minus(z1), minus(z0))
Tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
S tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS

Compound Symbols:

c1, c2

### (5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

minus(minus(z0)) → z0
minus(h(z0)) → h(minus(z0))
minus(f(z0, z1)) → f(minus(z1), minus(z0))

### (6) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
S tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:

MINUS

Compound Symbols:

c1, c2

### (7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
We considered the (Usable) Rules:none
And the Tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(MINUS(x1)) = [4] + [4]x1
POL(c1(x1)) = x1
POL(c2(x1, x2)) = x1 + x2
POL(f(x1, x2)) = [4] + x1 + x2
POL(h(x1)) = [5] + x1

### (8) Obligation:

Complexity Dependency Tuples Problem
Rules:none
Tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
S tuples:none
K tuples:

MINUS(h(z0)) → c1(MINUS(z0))
MINUS(f(z0, z1)) → c2(MINUS(z1), MINUS(z0))
Defined Rule Symbols:none

Defined Pair Symbols:

MINUS

Compound Symbols:

c1, c2

### (9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty